lecture13 - Titration A procedure where a base is added to...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Titration A procedure where a base is added to an acid, or an acid is added to a base until the stoichiometric amount of base or acid is added. This is called the equivalence point . The equivalence point is detected by a color change in an indicator. Strong Acid/Strong Base Titration Examples: HCl + NaOH HNO 3 + KOH Equation: H + (aq) + OH - (aq) H 2 O( ) K = When a reaction has a high K (as in a titration reaction) first react to completion before doing any equilibrium calculations. K for above reaction = = 1 x 10 14 Assume we add NaOH (source of OH - ions) to a solution initially containing 0.10 mol HCl in 1.00 L of solution. Make a table and then a plot of pH vs. mol OH - added. H + (aq) + OH - (aq) H 2 O( ) K = = 10 14 (large) Showing a calculation in detail (before the equivalence point): Adding 0.090 mol OH - to the initial 0.10 mol H + , all in 1.00 L. We react to completion since K is large, then calculate the pH based on the excess [H + ] concentration H + (aq) + OH - (aq) H 2 O( ) Start 0.10 0.090 RTC -0.090 -0.090 RTC = “react to completion” 0.010 “0” Excess [H + ] = 0.010 M pH = -log (0.010) = 2.00 Showing another calculation in detail (after the equivalence point): Adding 0.11 mol OH - to the initial 0.10 mol H + , all in 1.00 L. We react to completion since K is large, then calculate the pH based on the excess [OH - ] concentration
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
H + (aq) + OH - (aq) H 2 O( ) Start 0.10 0.11 RTC -0.10 -0.10 RTC = “react to completion” “0” 0.01 Excess [OH - ] = 0.010 M pOH = -log (0.010) = 2.00 pH = 14.00 - 2.00 = 12.00 Start with 0.10 mol HCl (total volume of 1.00 Liters) mol OH - added pH 0 1.00 initial 0.10 M HCl 0.010 1.05 excess [H + ]: 0.10 - 0.010 = 0.090 M 0.050 1.30 excess [H + ]: 0.10 - 0.050 = 0.050 M 0.090 2.00 excess [H + ]: 0.10 - 0.090 = 0.010 M 0.099 3.00 excess [H + ]: 0.10 - 0.099 = 0.001 M 0.0999 4.00 excess [H + ]: 0.10 - 0.0999 = 0.0001 M 0.10 7.00 equivalence point, no excess [H + ] or [OH - ] 0.1001 10.00 excess [OH - ]: 0.1001 - 0.10 = 0.0001 M 0.101 11.00 excess [OH - ]: 0.101 - 0.10 = 0.001 M 0.110 12.00 excess [OH - ]: 0.11 - 0.10 = 0.01 M 0.150 12.70 excess [OH - ]: 0.15 - 0.10 = 0.05 M 0.200 13.00 excess [OH - ]: 0.20 - 0.10 = 0.10 M
Background image of page 2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH 0 0.05 0.1 0.15 0.2 0.25 mol NaOH added 0.10 mol HCl titrated with NaOH (1.00 Liter total) pH Note the sharp change of pH near the equivalence point. A tiny amount of NaOH added near the equivalence point changes the pH from 3 to 10. The equivalence point (pH 7) is the point of inflection in the titration curve. This is the point where the slope changes from increasing to decreasing. For simplicity, we showed the above calculations assuming a constant volume of 1.00 liter, as we added mol NaOH to neutralize the HCl. In a real laboratory situation, we would add a solution of NaOH to a solution of HCl, and the volume would change. Suppose we are titrating 50.0 mL of 0.20 M HNO
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/22/2008 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

Page1 / 22

lecture13 - Titration A procedure where a base is added to...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online